Frictional Forces

1
Frictional Forces

• Chapter 4, Section 4
• Pg. 141-149

2
Friction

• There are two types of frictional forces in the
  physical world that effect us.
• - Static/Kinetic Friction
• - Air Resistance

3
Air Resistance

• Is a resistance force that acts in the opposite
  direction of gravitational forces.

Help!!
v1 gt v2
4
Static and Kinetic Friction

• Forces that oppose motion between two surfaces
  that are touching each other.

Fr (non-moving) Static Friction (Fs)
Fr (moving) Kinetic Friction (Fk)
5
Frictional force depends on the size and mass of
the object moving across a surface.
The amount of friction occurring between an
object and the surface depends on the surface
type.
6
Tile Floor
Carpet
7
The value that expresses the dependence of
frictional forces on the surface type the object
is in contact with is called the coefficient of
friction (µ).
It is the ratio between the force of friction and
the normal force.
8
Coefficient of Kinetic Friction
µk Fk/Fn coefficient of kinetic friction
µs Fs,max/Fn coefficient of static friction
Ff µFn frictional force
9
Sample problem
While redecorating her apartment, Suzy slowly
pushes an 82 kg cabinet across a wooden dining
room floor, which resists the motion with a force
of friction of 320 N. What is the coefficient of
kinetic friction between the cabinet and the
floor?
m 82 kg
g 9.81 m/s²
µk Fk/Fn
w mg
µk 320 N/804 N
w (82 kg) (9.81 m/s²)
w Fn 804 N
µk 0.40
w 804 N
10
Sample Problem 2 (Pg. 146)
A student moves a box of books by attaching a
rope to the box by pulling with a force of 90 N
at an angle 30 to the horizontal. The box of
books has a mass of 20.0 kg and the coefficient
of kinetic friction between the bottom of the box
and the sidewalk is 0.50. What is the
acceleration of the box?
11
Step 1 Solve for Weight and X Y components of
applied force
g 9.81 m/s²
m 20.0 kg
µk 0.50
30
w mg 196 N
Fapp, y (90N) (sin 30) 45.0 N
Fapp, x (90N) (cos 30) 77.9 N
12
Step 2 Find the Kinetic Friction
Fapp, y 45.0 N
Fapp, x 77.9 N
µk 0.50
w 196 N
?Fy Fn w Fapp, y 0
0 Fn 196 N 45.0 N
Fn 196 N 45 N 151 N
µk Fk/Fn
Fk (µk)(Fn) (0.50)(151 N) 75.5 N
13
Step 3 Solve for acceleration
m 20.0 kg
Fk 75.5 N
?F ma
Fapp, x 77.9 N
a ?F/m
a (Fapp, x Fk) / m
a (77.9 N 75.5 N) / 20.0 kg
a 0.12 m/s² to the right